A113319 -id:A113319 - cp's OEIS frontend

A002522 a(n) = n^2 + 1.

1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501

Offset: 0

N. J. A. Sloane

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).
a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.
As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001
a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003
Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006
Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.
The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005
Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov
1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006
For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007
Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007
{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010
For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010
The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010
n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.
a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011
Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011
a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011
Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011
If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014
The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014
a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016
Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016
For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017
Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018
a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019
a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022
From Klaus Purath, Apr 03 2025: (Start)
The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.
It follows from the above that this sequence is a subsequence of A031396. (End)

			G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 17*x^4 + 26*x^5 + 37*x^6 + 50*x^7 + 65*x^8 + ...

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
E. Gura and M. Maschler, Insights into Game Theory: An Alternative Mathematical Experience, Cambridge, 2008; p. 26.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, New York, 2001.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000. Format corrected by _Peter Kagey_, Jan 25 2016
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, On the Salient Regularities of Strings of Assembly Theory, Preprints (2024). See p. 19.
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints.org (2025). See p. 28.
R. P. Boas and N. J. A. Sloane, Correspondence, 1974
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 44, 56.
Giulio Cerbai and Luca Ferrari, Permutation patterns in genome rearrangement problems: the reversal model, arXiv:1903.08774 [math.CO], 2019. See p. 19.
S. Chaiken et al., Nonattacking Queens in a Rectangular Strip, arXiv:1105.5087 [math.CO], 2011.
Bryan Ek, Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics, arXiv:1804.05933 [math.CO], 2018.
R. M. Green and Tianyuan Xu, 2-roots for simply laced Weyl groups, arXiv:2204.09765 [math.RT], 2022.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv:1410.2657 [math.CO], 2014.
C. Homberger and V. Vatter, On the effective and automatic enumeration of polynomial permutation classes, arXiv:1308.4946 [math.CO], 2013.
L. B. W. Jolley, Summation of Series, Dover, 1961, p. 176.
S. J. Leon, Linear Algebra with Applications: the Perron-Frobenius Theorem [Cached copy at the Wayback Machine]
T. Mansour and J. West, Avoiding 2-letter signed patterns, arXiv:math/0207204 [math.CO], 2002.
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv:1508.07894 [math.NT], 2015.
Eric Weisstein's World of Mathematics, Number Picking
Eric Weisstein's World of Mathematics, Near-Square Prime
Helmut Wielandt, Unzerlegbare, nicht negative Matrizen, Math. Z. 52 (1950), 642-648. volume 52
Reinhard Zumkeller, Enumerations of Divisors
Index to values of cyclotomic polynomials of integer argument
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Left edge of A055096.
Cf. A059100, A117950, A087475, A117951, A114949, A117619 (sequences of form n^2 + K).
a(n+1) = A101220(n, n+1, 3).
Cf. A059592, A124808, A132411, A132414, A028872, A005408, A000124, A016813, A086514, A000125, A058331, A080856, A000127, A161701-A161704, A161706, A161707, A161708, A161710-A161713, A161715, A006261.
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), this sequence (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A002496 (primes).
Cf. A254858.
Cf. A302612, A302644, A302645, A302646.
Subsequence of A031396.

a002522 = (+ 1) . (^ 2)
a002522_list = scanl (+) 1 [1,3..]
-- Reinhard Zumkeller, Apr 06 2012

[n^2 + 1: n in [0..50]]; // Vincenzo Librandi, May 01 2011

A002522 := proc(n)
        numtheory[cyclotomic](4,n) ;
end proc:
seq(A002522(n),n=0..20) ; # R. J. Mathar, Feb 07 2014

Table[n^2 + 1, {n, 0, 50}]; (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)

A002522(n):=n^2+1$ makelist(A002522(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

a(n)=n^2+1 \\ Charles R Greathouse IV, Jun 10 2011

O.g.f.: (1-x+2*x^2)/((1-x)^3). - Eric Werley, Jun 27 2011
Sequences of the form a(n) = n^2 + K with offset 0 have o.g.f. (K - 2*K*x + K*x^2 + x + x^2)/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a*(n-3). - R. J. Mathar, Apr 28 2008
For n > 0: a(n-1) = A143053(A000290(n)) - 1. - Reinhard Zumkeller, Jul 20 2008
A143053(a(n)) = A000290(n+1). - Reinhard Zumkeller, Jul 20 2008
a(n)*a(n-2) = (n-1)^4 + 4. - Reinhard Zumkeller, Feb 12 2009
a(n) = A156798(n)/A087475(n). - Reinhard Zumkeller, Feb 16 2009
From Reinhard Zumkeller, Mar 08 2010: (Start)
a(n) = A170949(A002061(n+1));
A170949(a(n)) = A132411(n+1);
A170950(a(n)) = A002061(n+1). (End)
For n > 1, a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n - 2)^2 + (a(n) + n - 1 + a(n) + n)^2 = (n+1) *(6*n^4 + 18*n^3 + 26*n^2 + 19*n + 6) / 6 = (a(n) + n)^2 + ... + (a(n) + 2*n)^2. - Charlie Marion, Jan 10 2011
From Eric Werley, Jun 27 2011: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2.
a(n) = a(n-1) + 2*n - 1. (End)
a(n) = (n-1)^2 + 2(n-1) + 2 = 122 read in base n-1 (for n > 3). - Jason Kimberley, Oct 20 2011
a(n)*a(n+1) = a(n*(n+1) + 1) so a(1)*a(2) = a(3). More generally, a(n)*a(n+k) = a(n*(n+k) + 1) + k^2 - 1. - Jon Perry, Aug 01 2012
a(n) = (n!)^2* [x^n] BesselI(0, 2*sqrt(x))*(1+x). - Peter Luschny, Aug 25 2012
a(n) = A070216(n,1) for n > 0. - Reinhard Zumkeller, Nov 11 2012
E.g.f.: exp(x)*(1 + x + x^2). - Geoffrey Critzer, Aug 30 2013
a(n) = A254858(n-2,3) for n > 2. - Reinhard Zumkeller, Feb 09 2015
Sum_{n>=0} (-1)^n / a(n) = (1+Pi/sinh(Pi))/2 = 0.636014527491... = A367976 . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} 1/a(n) = (1 + Pi*coth(Pi))/2 = 2.076674... = A113319. - Vaclav Kotesovec, Apr 10 2016
4*a(n) = A001105(n-1) + A001105(n+1). - Bruno Berselli, Jul 03 2017
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi)*sinh(sqrt(2)*Pi).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi). (End)
Sum_{n>=0} a(n)/n! = 3*e. - Davide Rotondo, Feb 16 2025

Partially edited by Joerg Arndt, Mar 11 2010

A072691 Decimal expansion of Pi^2/12.

Original entry on oeis.org

8, 2, 2, 4, 6, 7, 0, 3, 3, 4, 2, 4, 1, 1, 3, 2, 1, 8, 2, 3, 6, 2, 0, 7, 5, 8, 3, 3, 2, 3, 0, 1, 2, 5, 9, 4, 6, 0, 9, 4, 7, 4, 9, 5, 0, 6, 0, 3, 3, 9, 9, 2, 1, 8, 8, 6, 7, 7, 7, 9, 1, 1, 4, 6, 8, 5, 0, 0, 3, 7, 3, 5, 2, 0, 1, 6, 0, 0, 4, 3, 6, 9, 1, 6, 8, 1, 4, 4, 5, 0, 3, 0, 9, 8, 7, 9, 3, 5, 2, 6, 5, 2, 0, 0, 2

Offset: 0

Rick L. Shepherd, Jul 02 2002

			0.822467033424113218236207583323..

C. C. Clawson, The Beauty and Magic of Numbers. New York: Plenum Press (1996): 98
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.11 p. 126 and section 8.5 p. 501.
Jolley, Summation of Series, Dover (1961) eq. (234) page 44.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Paul Bracken, Problem 4826, Crux Mathematicorum, Vol. 49, No. 3 (March, 2023), p. 157; Michel Bataille, Solution to Problem 4826, ibid., Vol. 49, No. 8 (Oct. 2023), p. 452.
Eugène-Charles Catalan, Mémoire sur la transformation des séries et sur quelques intégrales définies, Mémoires de l'Académie royale de Belgique, 1867, Vol. 33, pp. 1-50.
Brian Hawthorn, The Hardest Integral I've Ever Done, YouTube video, 2021.
Michael Penn, A viewer suggested integral, YouTube video, 2021.
Eric Weisstein's World of Mathematics, Dilogarithm
Index entries for transcendental numbers

Cf. A072692 (Pi^2/12 is in asymptotic formula related to sigma(n), A000203).
Cf. A113319 (sum_{i>=0} 1/(i^2+1)); A232883 (sum_{i>=0} 1/(2*i^2+1)).
Cf. A001008, A002805, A013661, A024916, A237593, A244583. Inverse of A377363.

RealDigits[Pi^2/12, 10, 105][[1]] (* Robert G. Wilson v *)

zeta(2)/2 \\ Michel Marcus, Sep 08 2014

-dilog(-1) \\ Charles R Greathouse IV, Apr 17 2015

Pi^2/12 \\ Charles R Greathouse IV, Apr 17 2015

sumnumrat(1/(2*x^2), 0) \\ Charles R Greathouse IV, Jan 20 2022

from mpmath import *
mp.dps=106
print([int(c) for c in list(str(zeta(2)/2))[2:-1]]) # Indranil Ghosh, Jul 08 2017

Equals 1/(1*2) + 1/(2*4) + 1/(3*6) + 1/(4*8) + ... [Jolley]
Equals -dilogarithm(-1). - Rick L. Shepherd, Jul 21 2004
Equals Sum_{n>=1} ((-1)^(n+1))/n^2 [Clawson]. - Alonso del Arte, Aug 15 2012
Equals Integral_{x=0..1} log((1+x^3)/(1-x^3))/x dx. - Bruno Berselli, May 13 2013
From Jean-François Alcover, May 17 2013: (Start)
Equals zeta(2)/2 = A013661/2.
Equals Integral_{x=1..2} log(x)/(x-1) dx. (End)
Equals lim_{n->infinity} A244583(n)/prime(n)^2. See A244583 for details. - Richard R. Forberg, Jan 04 2015
Equals Sum_{k>=1} H(k)/(k*2^k), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - Amiram Eldar, Aug 20 2020
Equals Integral_{0..infinity} x/(exp(x) + 1) dx. See Abramowitz-Stegun, 23.2.8, for s=2, p. 801. - Wolfdieter Lang, Sep 16 2020
Equals lim_{n->infinity} A024916(n)/(n^2). - Omar E. Pol, Dec 15 2021
Integral_{x=0..1} -log(x)/(x+1) dx. - Bernard Schott, Apr 25 2022
Equals 1/2 + Sum_{k>=1} H(k)/(k*(k+1)*(k+2)), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number (Bracken, 2023). - Amiram Eldar, Oct 06 2023
Equals Integral_{x >= 0} x^2/cosh(x)^2 dx. - Peter Bala, Jun 20 2024
Equals 1 + (1/8)*Sum_{k >= 0} (-1)^(k-1) * (10*k + 13)/((k + 1)*(2*k + 1)^2*(2*k + 3)^2*binomial(2*k, k)). See Catalan, Section 35, equation 54. - Peter Bala, Aug 17 2024
Equals Integral_{x=0..oo} ((arctan(x) - Pi/4)*log(x^2 + 1))/(x^2) dx. - Kritsada Moomuang, Jun 04 2025

A259171 Decimal expansion of Sum_{m>=1} 1/(m^2 + 1).

Original entry on oeis.org

1, 0, 7, 6, 6, 7, 4, 0, 4, 7, 4, 6, 8, 5, 8, 1, 1, 7, 4, 1, 3, 4, 0, 5, 0, 7, 9, 4, 7, 5, 0, 0, 0, 0, 4, 9, 0, 4, 4, 5, 6, 5, 6, 2, 6, 6, 4, 0, 3, 8, 1, 6, 6, 6, 5, 5, 7, 5, 0, 6, 2, 4, 8, 4, 3, 9, 0, 1, 5, 4, 2, 4, 7, 9, 1, 8, 3, 1, 0, 0, 2, 1, 7, 4, 3, 5

Offset: 1

José María Grau Ribas, Jun 19 2015

Essentially the same as A100554 and A113319. - R. J. Mathar, Jul 06 2015

			1.07667404746858117413405079475000049044565626640381666557

Muniru A Asiru, Table of n, a(n) for n = 1..3000
Junyong Zhao, Shaofang Hong, and Xiao Jiang, A certain reciprocal power sum is never an integer, arXiv:1812.08705 [math.NT], 2018. See the constant alpha_f.

Cf. A013661, A259173.

evalf[120]((Pi*coth(Pi)-1)/2); # Muniru A Asiru, Dec 21 2018

N[Sum[1/(k^2+1),{k,Infinity}],1000]//RealDigits//First

(Pi*cosh(Pi)/sinh(Pi)-1)/2 \\ Michel Marcus, Jun 28 2015

sumnumrat(1/(x^2+1), 1) \\ Charles R Greathouse IV, Jan 20 2022

Equals (Pi*coth(Pi)-1)/2. - Vaclav Kotesovec, Jun 27 2015
Equals Integral_{x>=0} sin(x)/(exp(x) - 1) dx. - Amiram Eldar, Aug 16 2020
Equals Integral_{x>=0} (sin(x)/sinh(x))^2 dx. - Amiram Eldar, Dec 11 2023

A329083 Decimal expansion of Sum_{k>=0} 1/(k^2+2).

Original entry on oeis.org

1, 3, 6, 1, 0, 2, 8, 1, 0, 0, 5, 7, 3, 7, 2, 7, 9, 2, 2, 8, 2, 1, 3, 3, 2, 1, 5, 8, 5, 1, 8, 2, 3, 4, 6, 3, 6, 8, 7, 2, 8, 5, 3, 5, 6, 0, 7, 0, 6, 9, 3, 0, 7, 2, 3, 3, 4, 9, 4, 7, 8, 9, 0, 0, 1, 6, 0, 7, 8, 2, 1, 1, 4, 6, 3, 6, 5, 5, 4, 4, 4, 5, 7, 3, 7, 6, 1, 5, 1, 4, 7

Offset: 1

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(2).
This and A329090 are essentially the same, but both sequences are added because some people may search for this, and some people may search for A329090.

			1.36102810057372792282...

Cf. A329080 (F(-5)), A329081 (F(-3)), A329082 (F(-2)), A113319 (F(1)), this sequence (F(2)), A329084 (F(3)), A329085 (F(4)), A329086 (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[(1 + Sqrt[2]*Pi*Coth[Sqrt[2]*Pi])/4, 10, 120][[1]] (* Amiram Eldar, Jun 17 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(2)

sumnumrat(1/(x^2+2),0) \\ Charles R Greathouse IV, Jan 20 2022

Equals (1 + (sqrt(2)*Pi)*coth(sqrt(2)*Pi))/4 = (1 + (sqrt(-2)*Pi)*cot(sqrt(-2)*Pi))/4.

A329084 Decimal expansion of Sum_{k>=0} 1/(k^2+3).

Original entry on oeis.org

1, 0, 7, 3, 6, 0, 0, 4, 0, 9, 9, 1, 5, 1, 8, 4, 1, 1, 5, 9, 1, 3, 9, 3, 6, 2, 9, 8, 1, 5, 8, 1, 4, 5, 3, 1, 1, 2, 7, 6, 4, 4, 2, 6, 3, 5, 7, 1, 8, 7, 8, 4, 5, 7, 8, 9, 6, 0, 3, 6, 8, 7, 5, 1, 9, 5, 8, 6, 6, 7, 5, 2, 3, 1, 8, 4, 5, 6, 3, 4, 5, 9, 8, 8, 5, 8, 4, 8, 2, 4, 9

Offset: 1

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(3).

			1.07360040991518411591...

Cf. A329080 (F(-5)), A329081 (F(-3)), A329082 (F(-2)), A113319 (F(1)), A329083 (F(2)), this sequence (F(3)), A329085 (F(4)), A329086 (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[Sum[1/(k^2+3),{k,0,\[Infinity]}],10,120][[1]] (* Harvey P. Dale, Jul 05 2021 *)
RealDigits[(1 + Sqrt[3]*Pi*Coth[Sqrt[3]*Pi])/6, 10, 120][[1]] (* Amiram Eldar, Jun 17 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(3)

sumnumrat(1/(x^2+3), 0) \\ Charles R Greathouse IV, Jan 20 2022

Equals (1 + (sqrt(3)*Pi)*coth(sqrt(3)*Pi))/6 = (1 + (sqrt(-3)*Pi)*cot(sqrt(-3)*Pi))/6.

A329085 Decimal expansion of Sum_{k>=0} 1/(k^2+4).

Original entry on oeis.org

9, 1, 0, 4, 0, 3, 6, 4, 1, 3, 2, 1, 1, 1, 5, 1, 1, 4, 1, 9, 3, 0, 4, 3, 8, 2, 4, 9, 2, 6, 4, 4, 3, 6, 0, 9, 6, 1, 1, 6, 9, 5, 0, 6, 5, 7, 9, 4, 6, 5, 0, 4, 4, 8, 9, 0, 2, 5, 8, 5, 8, 8, 0, 4, 5, 3, 5, 8, 0, 8, 3, 1, 1, 4, 9, 4, 5, 5, 2, 0, 6, 2, 5, 2, 8, 4, 5, 3, 1, 7, 8

Offset: 0

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(4).
This and A329092 are essentially the same, but both sequences are added because some people may search for this, and some people may search for A329092.

			0.91040364132111511419...

Cf. A329080 (F(-5)), A329081 (F(-3)), A329082 (F(-2)), A113319 (F(1)), A329083 (F(2)), A329084 (F(3)), this sequence (F(4)), A329086 (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[(1 + 2*Pi*Coth[2*Pi])/8, 10, 120][[1]] (* Amiram Eldar, Jun 17 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(4)

sumnumrat(1/(x^2+4), 0) \\ Charles R Greathouse IV, Jan 20 2022

Equals (1 + (2*Pi)*coth(2*Pi))/8 = (1 + (2*Pi*i)*cot(2*Pi*i))/8, i = sqrt(-1).

A329080 Decimal expansion of Sum_{k>=0} 1/(k^2-5), negated.

Original entry on oeis.org

8, 6, 6, 8, 3, 2, 5, 9, 5, 6, 6, 2, 7, 4, 4, 8, 5, 2, 9, 8, 2, 9, 6, 3, 3, 3, 9, 7, 6, 6, 9, 6, 8, 1, 5, 7, 5, 4, 3, 4, 3, 2, 5, 6, 6, 2, 3, 8, 0, 3, 9, 6, 4, 0, 4, 0, 5, 8, 3, 3, 4, 5, 8, 2, 7, 1, 4, 8, 6, 8, 3, 3, 7, 2, 8, 9, 9, 0, 6, 0, 3, 4, 3, 6, 8, 6, 0, 4, 9, 2, 1

Offset: 0

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(-5) (negated).
This and A329087 are essentially the same, but both sequences are added because some people may search for this, and some people may search for A329087.

			-0.86683259566274485298...

Cf. this sequence (F(-5)), A329081 (F(-3)), A329082 (F(-2)), A113319 (F(1)), A329083 (F(2)), A329084 (F(3)), A329085 (F(4)), A329086 (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[(1 + Sqrt[5]*Pi*Cot[Sqrt[5]*Pi])/10, 10, 120][[1]] (* Amiram Eldar, Jun 17 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(-5)

sumnumrat(1/(x^2-5), 0) \\ Charles R Greathouse IV, Jan 20 2022

Equals (1 + (sqrt(-5)*Pi)*coth(sqrt(-5)*Pi))/(-10).

Equals (1 + (sqrt(5)*Pi)*cot(sqrt(5)*Pi))/(-10).

A329081 Decimal expansion of Sum_{k>=0} 1/(k^2-3).

Original entry on oeis.org

6, 4, 3, 3, 1, 6, 8, 5, 6, 6, 5, 2, 7, 6, 0, 2, 8, 3, 7, 7, 2, 5, 1, 5, 7, 2, 1, 8, 0, 8, 3, 8, 2, 9, 2, 9, 1, 0, 9, 7, 2, 6, 0, 7, 8, 1, 1, 2, 1, 8, 3, 5, 8, 6, 0, 5, 3, 6, 3, 2, 8, 4, 3, 3, 0, 1, 8, 8, 3, 1, 8, 4, 9, 5, 8, 3, 9, 6, 0, 3, 6, 9, 2, 6, 1, 4, 7, 1, 1, 8, 8

Offset: 0

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(-3).

			0.64331685665276028377...

Cf. A329080 (F(-5)), this sequence (F(-3)), A329082 (F(-2)), A113319 (F(1)), A329083 (F(2)), A329084 (F(3)), A329085 (F(4)), A329086 (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[(1 + Sqrt[3]*Pi*Cot[Sqrt[3]*Pi])/6, 10, 120][[1]] (* Amiram Eldar, Jun 17 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(-3)

sumnumrat(1/(x^2-3), 0) \\ Charles R Greathouse IV, Jan 20 2022

Equals (1 + (sqrt(-3)*Pi)*coth(sqrt(-3)*Pi))/(-6) = (1 + (sqrt(3)*Pi)*cot(sqrt(3)*Pi))/(-6).

A329082 Decimal expansion of Sum_{k>=0} 1/(k^2-2), negated.

Original entry on oeis.org

5, 5, 6, 8, 1, 0, 4, 0, 7, 7, 0, 0, 6, 2, 0, 0, 8, 2, 5, 5, 2, 9, 8, 1, 6, 0, 9, 1, 1, 2, 5, 9, 7, 3, 4, 7, 0, 9, 8, 7, 0, 9, 2, 7, 0, 2, 5, 7, 0, 4, 0, 8, 7, 8, 5, 5, 1, 0, 0, 1, 9, 8, 3, 4, 8, 6, 3, 2, 8, 1, 0, 3, 7, 4, 4, 1, 5, 7, 0, 0, 2, 4, 6, 1, 7, 4, 5, 6, 5, 7, 7

Offset: 0

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(-2) (negated).
This and A329089 are essentially the same, but both sequences are added because some people may search for this, and some people may search for A329089.

			-0.55681040770062008255...

Cf. A329080 (F(-5)), A329081 (F(-3)), this sequence (F(-2)), A113319 (F(1)), A329083 (F(2)), A329084 (F(3)), A329085 (F(4)), A329086 (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[(1 + Sqrt[2]*Pi*Cot[Sqrt[2]*Pi])/4, 10, 120][[1]] (* Amiram Eldar, Jun 17 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(-2)

sumnumrat(1/(x^2-2), 0) \\ Charles R Greathouse IV, Jan 20 2022

Equals (1 + (sqrt(-2)*Pi)*coth(sqrt(-2)*Pi))/(-4) = (1 + (sqrt(2)*Pi)*cot(sqrt(2)*Pi))/(-4).

A329086 Decimal expansion of Sum_{k>=0} 1/(k^2+5).

Original entry on oeis.org

8, 0, 2, 4, 8, 2, 5, 8, 4, 8, 0, 6, 7, 8, 6, 8, 8, 6, 8, 3, 5, 8, 4, 4, 9, 5, 4, 4, 8, 6, 5, 5, 7, 7, 0, 9, 4, 0, 7, 1, 6, 0, 7, 2, 9, 7, 9, 0, 5, 7, 0, 1, 3, 6, 4, 1, 9, 8, 5, 9, 5, 9, 3, 9, 6, 0, 9, 4, 0, 1, 4, 9, 5, 4, 0, 5, 3, 4, 0, 8, 0, 4, 5, 5, 2, 9, 1, 0, 9, 3, 9

Offset: 0

Jianing Song, Nov 04 2019

In general, for complex numbers z, if we define F(z) = Sum_{k>=0} 1/(k^2+z), f(z) = Sum_{k>=1} 1/(k^2+z), then we have:
F(z) = (1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...;
f(z) = (-1 + sqrt(z)*Pi*coth(sqrt(z)*Pi))/(2z), z != 0, -1, -4, -9, -16, ...; Pi^2/6, z = 0. Note that f(z) is continuous at z = 0.
This sequence gives F(5).
This and A329093 are essentially the same, but both sequences are added because some people may search for this, and some people may search for A329093.

			Sum_{k>=0} 1/(k^2+5) = 0.80248258480678688683...

Cf. A329080 (F(-5)), A329081 (F(-3)), A329082 (F(-2)), A113319 (F(1)), A329083 (F(2)), A329084 (F(3)), A329085 (F(4)), this sequence (F(5)).

Cf. A329087 (f(-5)), A329088 (f(-3)), A329089 (f(-2)), A013661 (f(0)), A259171 (f(1)), A329090 (f(2)), A329091 (f(3)), A329092 (f(4)), A329093 (f(5)).

RealDigits[(1 + Sqrt[5]*Pi*Coth[Sqrt[5]*Pi])/10, 10, 120][[1]] (* Amiram Eldar, Jun 15 2023 *)

default(realprecision, 100); my(F(x) = (1 + (sqrt(x)*Pi)/tanh(sqrt(x)*Pi))/(2*x)); F(5)

sumnumrat(1/(x^2+5), 0) \\ Charles R Greathouse IV, Jan 20 2022

Sum_{k>=0} 1/(k^2+5) = (1 + (sqrt(5)*Pi)*coth(sqrt(5)*Pi))/10 = (1 + (sqrt(-5)*Pi)*cot(sqrt(-5)*Pi))/10.

cp's OEIS Frontend

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